Boyce/Diprima 9 Ed, CH 2.2: Separable Equations: Y X F DX Dy
Boyce/Diprima 9 Ed, CH 2.2: Separable Equations: Y X F DX Dy
Boyce/Diprima 9 Ed, CH 2.2: Separable Equations: Y X F DX Dy
2:
Separable Equations
Elementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
∫ (1 − y )dy = ∫ (x )dx
2 2 2
1 1 -4 -2 2 4
xH t L
y − y3 = x3 + C
3 3 -2
3 y − y 3 = x3 + C -4
(
y − 2 y − x + 2x + 2x + C = 0 ⇒ y =
2 3 2
)2 ± 4 + 4 x 3
(+ 2 x 2
+ 2x + C )
2
y = 1 ± x3 + 2x 2 + 2x + C
dy 3 x 2 + 4 x + 2
=
dx 2( y − 1)
y = 1 + x3 + 2 x 2 + 2 x + 4
Example 2: Domain (4 of 4)
Thus the solutions to the initial value problem
dy 3 x 2 + 4 x + 2
= , y (0) = −1
dx 2( y − 1)
are given by
y 2 − 2 y = x 3 + 2 x 2 + 2 x + 3 (implicit)
y = 1 − x3 + 2 x 2 + 2 x + 4 (explicit)
From explicit representation of y, it follows that
( )
y = 1 − x 2 (x + 2 ) + 2(x + 2) = 1 − (x + 2) x 2 + 2
and hence the domain of y is (-2, ∞). Note x = -2 yields y = 1, which
makes the denominator of dy/dx zero (vertical tangent).
Conversely, the domain of y can be estimated by locating vertical
tangents on the graph (useful for implicitly defined solutions).
Example 3: Implicit Solution of an Initial
Value Problem (1 of 2)
Consider the following initial value problem:
4 x − x3
y′ = 3
, y ( 0) = 1
4+ y
Separating variables and using calculus, we obtain
(4 + y 3 )dy = (4 x − x 3 )dx
∫ (4 + y )dy = ∫ (4 x − x )dx
3 3
1 1
4 y + y4 = 2x2 − x4 + c
4 4
y 4 + 16 y + x 4 − 8 x 2 = C where C = 4c
Using the initial condition, y(0)=1, it follows that C = 17.
y 4 + 16 y + x 4 − 8 x 2 = 17
4 x − x3
y′ = , y (0) = 1
4 + y3